Question: In a circle with center $O$, $AD$ is a diameter, $ABC$ is a chord, $BO = 5$, and $\angle ABO = \text{arc } CD = 60^\circ$.  Find the length of $BC$.

[asy]
import graph;

unitsize(2 cm);

pair O, A, B, C, D;

O = (0,0);
A = dir(30);
C = dir(160);
B = (2*C + A)/3;
D = -A;

draw(Circle(O,1));
draw(C--A--D);
draw(B--O);

label("$A$", A, NE);
label("$B$", B, N);
label("$C$", C, W);
label("$D$", D, SW);
label("$O$", O, SE);
[/asy]
Explanation: Since arc $CD$ is $60^\circ$, $\angle CAD = 60^\circ/2 = 30^\circ$.  Since triangle $AOC$ is isosceles with $AO = CO$, $\angle OCA = \angle OAC = 30^\circ$.

[asy]
import graph;

unitsize(2 cm);
pair O, A, B, C, D;

O = (0,0);
A = dir(30);
C = dir(160);
B = (2*C + A)/3;
D = -A;

draw(Circle(O,1));
draw(C--A--D);
draw(B--O);
draw(C--O);

label("$A$", A, NE);
label("$B$", B, N);
label("$C$", C, W);
label("$D$", D, SW);
label("$O$", O, SE);
[/asy]

Since $\angle ABO = 60^\circ$, and this angle is external to triangle $BCO$, $\angle BOC = \angle ABO - \angle BCO = 60^\circ - 30^\circ = 30^\circ$.  Hence, triangle $BCO$ is isosceles, and $BC = BO = \boxed{5}$.